mirror of https://github.com/django/django.git
275 lines
8.6 KiB
Plaintext
275 lines
8.6 KiB
Plaintext
==============
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Managing files
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==============
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This document describes Django's file access APIs for files such as those
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uploaded by a user. The lower level APIs are general enough that you could use
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them for other purposes. If you want to handle "static files" (JS, CSS, etc.),
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see :doc:`/howto/static-files/index`.
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By default, Django stores files locally, using the :setting:`MEDIA_ROOT` and
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:setting:`MEDIA_URL` settings. The examples below assume that you're using these
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defaults.
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However, Django provides ways to write custom `file storage systems`_ that
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allow you to completely customize where and how Django stores files. The
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second half of this document describes how these storage systems work.
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.. _file storage systems: `File storage`_
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Using files in models
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=====================
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When you use a :class:`~django.db.models.FileField` or
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:class:`~django.db.models.ImageField`, Django provides a set of APIs you can use
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to deal with that file.
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Consider the following model, using an :class:`~django.db.models.ImageField` to
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store a photo::
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from django.db import models
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class Car(models.Model):
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name = models.CharField(max_length=255)
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price = models.DecimalField(max_digits=5, decimal_places=2)
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photo = models.ImageField(upload_to="cars")
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specs = models.FileField(upload_to="specs")
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Any ``Car`` instance will have a ``photo`` attribute that you can use to get at
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the details of the attached photo:
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.. code-block:: pycon
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>>> car = Car.objects.get(name="57 Chevy")
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>>> car.photo
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<ImageFieldFile: cars/chevy.jpg>
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>>> car.photo.name
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'cars/chevy.jpg'
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>>> car.photo.path
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'/media/cars/chevy.jpg'
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>>> car.photo.url
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'http://media.example.com/cars/chevy.jpg'
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This object -- ``car.photo`` in the example -- is a ``File`` object, which means
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it has all the methods and attributes described below.
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.. note::
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The file is saved as part of saving the model in the database, so the actual
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file name used on disk cannot be relied on until after the model has been
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saved.
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For example, you can change the file name by setting the file's
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:attr:`~django.core.files.File.name` to a path relative to the file storage's
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location (:setting:`MEDIA_ROOT` if you are using the default
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:class:`~django.core.files.storage.FileSystemStorage`):
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.. code-block:: pycon
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>>> import os
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>>> from django.conf import settings
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>>> initial_path = car.photo.path
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>>> car.photo.name = "cars/chevy_ii.jpg"
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>>> new_path = settings.MEDIA_ROOT + car.photo.name
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>>> # Move the file on the filesystem
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>>> os.rename(initial_path, new_path)
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>>> car.save()
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>>> car.photo.path
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'/media/cars/chevy_ii.jpg'
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>>> car.photo.path == new_path
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True
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To save an existing file on disk to a :class:`~django.db.models.FileField`:
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.. code-block:: pycon
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>>> from pathlib import Path
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>>> from django.core.files import File
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>>> path = Path("/some/external/specs.pdf")
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>>> car = Car.objects.get(name="57 Chevy")
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>>> with path.open(mode="rb") as f:
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... car.specs = File(f, name=path.name)
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... car.save()
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...
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.. note::
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While :class:`~django.db.models.ImageField` non-image data attributes, such
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as ``height``, ``width``, and ``size`` are available on the instance, the
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underlying image data cannot be used without reopening the image. For
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example:
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.. code-block:: pycon
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>>> from PIL import Image
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>>> car = Car.objects.get(name="57 Chevy")
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>>> car.photo.width
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191
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>>> car.photo.height
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287
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>>> image = Image.open(car.photo)
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# Raises ValueError: seek of closed file.
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>>> car.photo.open()
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<ImageFieldFile: cars/chevy.jpg>
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>>> image = Image.open(car.photo)
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>>> image
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<PIL.JpegImagePlugin.JpegImageFile image mode=RGB size=191x287 at 0x7F99A94E9048>
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The ``File`` object
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===================
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Internally, Django uses a :class:`django.core.files.File` instance any time it
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needs to represent a file.
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Most of the time you'll use a ``File`` that Django's given you (i.e. a file
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attached to a model as above, or perhaps an uploaded file).
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If you need to construct a ``File`` yourself, the easiest way is to create one
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using a Python built-in ``file`` object:
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.. code-block:: pycon
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>>> from django.core.files import File
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# Create a Python file object using open()
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>>> f = open("/path/to/hello.world", "w")
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>>> myfile = File(f)
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Now you can use any of the documented attributes and methods
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of the :class:`~django.core.files.File` class.
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Be aware that files created in this way are not automatically closed.
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The following approach may be used to close files automatically:
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.. code-block:: pycon
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>>> from django.core.files import File
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# Create a Python file object using open() and the with statement
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>>> with open("/path/to/hello.world", "w") as f:
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... myfile = File(f)
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... myfile.write("Hello World")
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...
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>>> myfile.closed
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True
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>>> f.closed
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True
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Closing files is especially important when accessing file fields in a loop
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over a large number of objects. If files are not manually closed after
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accessing them, the risk of running out of file descriptors may arise. This
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may lead to the following error:
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.. code-block:: pytb
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OSError: [Errno 24] Too many open files
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File storage
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============
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Behind the scenes, Django delegates decisions about how and where to store files
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to a file storage system. This is the object that actually understands things
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like file systems, opening and reading files, etc.
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Django's default file storage is
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``'``:class:`django.core.files.storage.FileSystemStorage`\ ``'``. If you don't
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explicitly provide a storage system in the ``default`` key of the
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:setting:`STORAGES` setting, this is the one that will be used.
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See below for details of the built-in default file storage system, and see
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:doc:`/howto/custom-file-storage` for information on writing your own file
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storage system.
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Storage objects
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---------------
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Though most of the time you'll want to use a ``File`` object (which delegates to
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the proper storage for that file), you can use file storage systems directly.
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You can create an instance of some custom file storage class, or -- often more
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useful -- you can use the global default storage system:
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.. code-block:: pycon
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>>> from django.core.files.base import ContentFile
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>>> from django.core.files.storage import default_storage
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>>> path = default_storage.save("path/to/file", ContentFile(b"new content"))
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>>> path
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'path/to/file'
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>>> default_storage.size(path)
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11
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>>> default_storage.open(path).read()
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b'new content'
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>>> default_storage.delete(path)
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>>> default_storage.exists(path)
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False
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See :doc:`/ref/files/storage` for the file storage API.
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.. _builtin-fs-storage:
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The built-in filesystem storage class
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-------------------------------------
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Django ships with a :class:`django.core.files.storage.FileSystemStorage` class
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which implements basic local filesystem file storage.
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For example, the following code will store uploaded files under
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``/media/photos`` regardless of what your :setting:`MEDIA_ROOT` setting is::
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from django.core.files.storage import FileSystemStorage
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from django.db import models
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fs = FileSystemStorage(location="/media/photos")
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class Car(models.Model):
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...
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photo = models.ImageField(storage=fs)
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:doc:`Custom storage systems </howto/custom-file-storage>` work the same way:
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you can pass them in as the ``storage`` argument to a
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:class:`~django.db.models.FileField`.
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Using a callable
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----------------
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You can use a callable as the :attr:`~django.db.models.FileField.storage`
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parameter for :class:`~django.db.models.FileField` or
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:class:`~django.db.models.ImageField`. This allows you to modify the used
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storage at runtime, selecting different storages for different environments,
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for example.
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Your callable will be evaluated when your models classes are loaded, and must
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return an instance of :class:`~django.core.files.storage.Storage`.
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For example::
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from django.conf import settings
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from django.db import models
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from .storages import MyLocalStorage, MyRemoteStorage
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def select_storage():
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return MyLocalStorage() if settings.DEBUG else MyRemoteStorage()
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class MyModel(models.Model):
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my_file = models.FileField(storage=select_storage)
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In order to set a storage defined in the :setting:`STORAGES` setting you can
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use :data:`~django.core.files.storage.storages`::
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from django.core.files.storage import storages
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def select_storage():
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return storages["mystorage"]
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class MyModel(models.Model):
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upload = models.FileField(storage=select_storage)
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